Tuesday, November 27
Martin Kassabov, Cornell UniversityLet $G$ be a finitely generated residually finite group. Let $d(G)$ denote the minimal size of a generating set for $G$, and let $d(\widehat G)$ be the minimal size of a generating set for the profinite completion $\widehat G$ of $G$. In other words
d(\widehat G)= \max \left\{ d(G/H) \ | \ N \vartriangleleft G, \ G/N < \infty \right\}.
It is obvious that $d(G) \geq d(\widehat G)$ but there is no (easy) way to bound $d(G)$ from above -- there are examples of groups with abitrarly large $d(G)$ while $d(\widehat G)=2$.
Fortunately for polycyclic groups the situation is not that bad. Linnell and Warhurst prove that $d(G) \leq d(\widehat G) +1$ using methods from commutative algebra and lattices over orders.
In this talk I will give an alternative proof of this result. Moreover our result gives some sufficient condition when $d(G)=d(\widehat G)$ which can be verified quite easily in the case when $G$ is virtually abelian.
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